Internal problem ID [10122]
Internal file name [OUTPUT/9069_Monday_June_06_2022_06_21_28_AM_82955573/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1801 (book 6.210).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y \left (y^{2}+1\right ) y^{\prime \prime }+\left (1-3 y^{2}\right ) {y^{\prime }}^{2}=0} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} \left (y^{3}+y \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (-3 y^{2} p \left (y \right )+p \left (y \right )\right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {p \left (3 y^{2}-1\right )}{y \left (y^{2}+1\right )} \end {align*}
Where \(f(y)=\frac {3 y^{2}-1}{\left (y^{2}+1\right ) y}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {3 y^{2}-1}{\left (y^{2}+1\right ) y} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {\frac {3 y^{2}-1}{\left (y^{2}+1\right ) y} \,d y}\\ \ln \left (p \right )&=-\ln \left (y \right )+2 \ln \left (y^{2}+1\right )+c_{1}\\ p&={\mathrm e}^{-\ln \left (y \right )+2 \ln \left (y^{2}+1\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-\ln \left (y \right )+2 \ln \left (y^{2}+1\right )} \end {align*}
Which simplifies to \[ p \left (y \right ) = c_{1} \left (y^{3}+2 y +\frac {1}{y}\right ) \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = c_{1} \left (y^{3}+2 y+\frac {1}{y}\right ) \end {align*}
Integrating both sides gives \begin {align*} \int \frac {y}{c_{1} \left (y^{4}+2 y^{2}+1\right )}d y &= x +c_{2}\\ -\frac {1}{2 \left (y^{2}+1\right ) c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-\frac {\sqrt {-2 c_{1} \left (x +c_{2} \right ) \left (2 c_{1} c_{2} +2 c_{1} x +1\right )}}{2 c_{1} \left (x +c_{2} \right )}\\ y_2&=\frac {\sqrt {-2 c_{1} \left (x +c_{2} \right ) \left (2 c_{1} c_{2} +2 c_{1} x +1\right )}}{2 c_{1} \left (x +c_{2} \right )} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\sqrt {-2 c_{1} \left (x +c_{2} \right ) \left (2 c_{1} c_{2} +2 c_{1} x +1\right )}}{2 c_{1} \left (x +c_{2} \right )} \\ \tag{2} y &= \frac {\sqrt {-2 c_{1} \left (x +c_{2} \right ) \left (2 c_{1} c_{2} +2 c_{1} x +1\right )}}{2 c_{1} \left (x +c_{2} \right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\sqrt {-2 c_{1} \left (x +c_{2} \right ) \left (2 c_{1} c_{2} +2 c_{1} x +1\right )}}{2 c_{1} \left (x +c_{2} \right )} \] Verified OK.
\[ y = \frac {\sqrt {-2 c_{1} \left (x +c_{2} \right ) \left (2 c_{1} c_{2} +2 c_{1} x +1\right )}}{2 c_{1} \left (x +c_{2} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (y^{3}+y\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-3 y^{2} y^{\prime }+y^{\prime }\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),\frac {d}{d x}y^{\prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (y^{3}+y \right ) u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+\left (-3 y^{2} u \left (y \right )+u \left (y \right )\right ) u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {-3 y^{2} u \left (y \right )+u \left (y \right )}{y^{3}+y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=\frac {3 y^{2}-1}{\left (y^{2}+1\right ) y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}d y =\int \frac {3 y^{2}-1}{\left (y^{2}+1\right ) y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=-\ln \left (y \right )+2 \ln \left (y^{2}+1\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {{\mathrm e}^{c_{1}} \left (y^{4}+2 y^{2}+1\right )}{y} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {{\mathrm e}^{c_{1}} \left (y^{4}+2 y^{2}+1\right )}{y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{c_{1}} \left (y^{4}+2 y^{2}+1\right )}{y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{c_{1}} \left (y^{4}+2 y^{2}+1\right )}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{y^{4}+2 y^{2}+1}={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{y^{4}+2 y^{2}+1}d x =\int {\mathrm e}^{c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 \left (y^{2}+1\right )}={\mathrm e}^{c_{1}} x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {-2 \left ({\mathrm e}^{c_{1}} x +c_{2} \right ) \left (2 \,{\mathrm e}^{c_{1}} x +2 c_{2} +1\right )}}{2 \left ({\mathrm e}^{c_{1}} x +c_{2} \right )}, y=\frac {\sqrt {-2 \left ({\mathrm e}^{c_{1}} x +c_{2} \right ) \left (2 \,{\mathrm e}^{c_{1}} x +2 c_{2} +1\right )}}{2 \left ({\mathrm e}^{c_{1}} x +c_{2} \right )}\right \} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 27.406 (sec). Leaf size: 80
dsolve(y(x)*(y(x)^2+1)*diff(diff(y(x),x),x)+(1-3*y(x)^2)*diff(y(x),x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -i \\ y \left (x \right ) &= i \\ y \left (x \right ) &= 0 \\ y \left (x \right ) &= -\frac {\sqrt {2}\, \sqrt {-2 \left (c_{1} x +c_{2} +\frac {1}{2}\right ) \left (c_{1} x +c_{2} \right )}}{2 c_{1} x +2 c_{2}} \\ y \left (x \right ) &= \frac {\sqrt {2}\, \sqrt {-2 \left (c_{1} x +c_{2} +\frac {1}{2}\right ) \left (c_{1} x +c_{2} \right )}}{2 c_{1} x +2 c_{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 2.565 (sec). Leaf size: 223
DSolve[(1 - 3*y[x]^2)*y'[x]^2 + y[x]*(1 + y[x]^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {-2 c_1 x-1-2 c_2 c_1}}{\sqrt {2} \sqrt {c_1 (x+c_2)}} \\ y(x)\to \frac {\sqrt {-2 c_1 x-1-2 c_2 c_1}}{\sqrt {2} \sqrt {c_1 (x+c_2)}} \\ y(x)\to \text {Indeterminate} \\ y(x)\to \frac {\sqrt {-c_1}}{\sqrt {c_1}} \\ y(x)\to \frac {\sqrt {c_1}}{\sqrt {-c_1}} \\ y(x)\to -\frac {\sqrt {-x-c_2}}{\sqrt {x+c_2}} \\ y(x)\to \frac {\sqrt {-x-c_2}}{\sqrt {x+c_2}} \\ y(x)\to -\frac {\sqrt {x+c_2}}{\sqrt {-x-c_2}} \\ y(x)\to \frac {\sqrt {x+c_2}}{\sqrt {-x-c_2}} \\ \end{align*}